08-24-2007, 01:50 AM
its easy..
H-B equation we know rt
p2+2pq+q2=1
p2=frequency of homozygous domimant person,q2=frequency of homozygous recessive person,2pq=frequency of heterzygous person
so here given fequency is 1% which means p2=1 and p we get by square root of 1 i.e 0.1 got it till here???
and from second equation p+q=1 so we get q=0.1-1=0.9
so now hetrozygous frequency we get is 2pq=2X 0.1X0.9=0.18
so if he mates with hetrogygous female chance is 25 % they will get oofspring homozygous reccesive i.e= 0.25,therefore we multiply both 0.25 X 0.18=0.045
now its also possible that he mates with a homozygous female i.e q2=0.1 X 0.1=0.01 (because above we got value of q)
so if he does then chance sare 50% so chance is 0.01/2 =0.05
now adding 0.045 +0.005=0.05 is teh right answer.
(its kaplan physiology bank question)
H-B equation we know rt
p2+2pq+q2=1
p2=frequency of homozygous domimant person,q2=frequency of homozygous recessive person,2pq=frequency of heterzygous person
so here given fequency is 1% which means p2=1 and p we get by square root of 1 i.e 0.1 got it till here???
and from second equation p+q=1 so we get q=0.1-1=0.9
so now hetrozygous frequency we get is 2pq=2X 0.1X0.9=0.18
so if he mates with hetrogygous female chance is 25 % they will get oofspring homozygous reccesive i.e= 0.25,therefore we multiply both 0.25 X 0.18=0.045
now its also possible that he mates with a homozygous female i.e q2=0.1 X 0.1=0.01 (because above we got value of q)
so if he does then chance sare 50% so chance is 0.01/2 =0.05
now adding 0.045 +0.005=0.05 is teh right answer.
(its kaplan physiology bank question)